What Happened 12/

Class of 11/30 basically talked about different Theorem related to the Properties of Integral. Those Theorem are built around the Fundamental Theorem of Calculus which we will discuss in next class meeting.

 

We talked about Theorem 7.4.1 which states that Assume f : [a, b] \to \mathbb{R} is bounded, and let c \in (a, b). Then, f is integrable on [a, b] if and only if f is integrable on [a, c] and [c, b]. In this case, we have \int_a^b f = \int^c_a f + \int^b_c . The proof of this theorem uses the definition of refining a partition and the proving techniques of the Riemann integral.

 

Then we introduces Theorem 7.4.2 which catalogs the basic properties of integral.

Theorem 7.4.2 Assume f and g are integrable functions on the interval [a, b].

1). The function f + g is integrable on [a, b] with \int^b_a (f + g) = \int^b_a f + \int^b_a g.

2). For k \in \mathbb{R}, the function kf is integrable with \int^b_a kf = k \int^b_a f.

3).If m \leq f(x) \leq M on [a, b], then m(b − a) \leq \int^b_a f \leq M(b − a).

4). If f(x) \leq g(x) on [a, b], then \int^b_a f \leq \int^b_a g.

5.The function |f| is integrable and | \int^b_a f | \leq \int^b_a |f|.

 

The proof of (2) requires an extra attention that we need two cases where k can be positive or negative. If k is positive, then for any partition P we have U(kf,P) = kU(f,P) and L(kf,P) = kL(f,P). And since f is integrable, there exists partition (P_n) and consider function (kf). We have \lim\limits_{n \to \infty} [U(kf, P_n) − L(kf, P_n)] = \lim\limits_{n to \infty} k[U(f,P_n) − L(f,P_n)] = 0. And if k is negative then we have U(kf,P) = kU(f,P) and L(kf,P) = kL(f,P).

Posted on by Zehao Dong
Categories: What Happened

One Response

  1. Jeremy LeCrone says:
    December 4, 2017 at 10:08 am

    Thank you for the synopsis from class on Thursday.

    There is an important error in your last line: Namely, when k < 0, we actually have $latex U(kf,P) = k L(f,P)$ and $latex L(kf, P) = kU(f,P)$ for every partition. Recall that negating the values in a set effectively "flips" the set over so that supremum gets mapped to infimum and vice versa...

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